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ajax调用返回php接口返回json数据的方法(必看篇)
类别:PHP编程   作者:码皇   来源:互联网   点击:

下面小编就为大家带来一篇ajax调用返回php接口返回json数据的方法(必看篇)。小编觉得挺不错的,现在就分享给大家,也给大家做个参考。一起跟随小编过来看看吧

php代码如下:

    <?php header('Content-Type: application/json');
    header('Content-Type: text/html;
    charset=utf-8');
    $email = $_GET['email'];
    $user = [];
    $conn = @mysql_connect("localhost","Test","123456") or die("Failed in connecting database");
    mysql_select_db("Test",$conn);
    mysql_query("set names 'UTF-8'");
    $query = "select * from UserInformation where email = '".$email."'";
    $result = mysql_query($query);
    if (null == ($row = mysql_fetch_array($result))) {
    echo $_GET['callback']."(no such user)";
    }
    else {
    $user['email'] = $email;
    $user['nickname'] = $row['nickname'];
    $user['portrait'] = $row['portrait'];
    echo $_GET['callback']."(".json_encode($user).")";
    }
    ?>

js代码如下:

    <script> $.ajax({
    url: "http://test.localhost/UserInterfaceForChatroom/UserInformation.php?email=pshuyue@gmail.com", type: "GET", dataType: 'jsonp', // crossDomain: true, success: function (result) {
    // data = $.parseJSON(result);
    // alert(data.nickname);
    alert(result.nickname);
    }
    }
    );
    </script>

其中遇到了两个问题:

1、第一个问题:

Uncaught SyntaxError: Unexpected token :

解决方案如下:

This has just happened to me, and the reason was none of the reasons above. I was using the jQuery command getJSON and adding callback=? to use JSONP (as I needed to go cross-domain), and returning the JSON code {"foo":"bar"} and getting the error.

This is because I should have included the callback data, something like jQuery17209314005577471107_1335958194322({"foo":"bar"})

Here is the PHP code I used to achieve this, which degrades if JSON (without a callback) is used:

    $ret['foo'] = "bar";
    finish();
    function finish() {
    header("content-type:application/json");
    if ($_GET['callback']) {
    print $_GET['callback']."(";
    }
    print json_encode($GLOBALS['ret']);
    if ($_GET['callback']) {
    print ")";
    }
    exit;
    }

Hopefully that will help someone in the future.

2、第二个问题:

解析json数据。从上面的javascript中可以看到,我没有使用jquery.parseJSON()这些方法,开始使用这些方法,但是总是会报

VM219:1 Uncaught SyntaxError: Unexpected token o in JSON at position 1的错误,后来不用jquery.parseJSON()这个方法,反而一切正常。不知为何。

以上这篇ajax调用返回php接口返回json数据的方法(必看篇)就是小编分享给大家的全部内容了,希望能给大家一个参考,也希望大家多多支持脚本之家。

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